# If dy/dt = ky and k is a nonzero constant, y could be

by lude1
Tags: constant, dy or dt, nonzero
 P: 34 1. The problem statement, all variables and given/known data If dy/dt = ky and k is a nonzero constant, than y could be a. 2e^kty b. 2e^kt c. e^kt + 3 d. kty + 5 e. .5ky^2 + .5 Correct answer is b. 2e^kt 2. Relevant equations 3. The attempt at a solution I don't really know what I should be doing. I have a feeling I have to work backwards, therefore do the integral of ky? But if that is the case, I have no idea how e^kt is created through that integral. 2e^kt reminded me of Ce^kt, but that doesn't have anything to do with this problem either. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
 P: 58 Look at the derivative of e^kt and see how that relates to e^kt.
 HW Helper P: 6,202 The 'correct' thing to do would be dy/dt = ky ∫ dy/y = ∫ k dt But since you have a multiple choice, you could quickly differentiate each one if you wanted to.
 P: 34 If dy/dt = ky and k is a nonzero constant, y could be Hm.. this is what I got when I did ∫ dy/y = ∫ k dt lny = kt y = e^kt Where does the 2 come from? I guessed it would have to be from a square, but when you do the integral of a number (since k is a nonzero constant) you multiply the constant with a variable with an exponent of 1, not 2.
HW Helper
P: 6,202
 Quote by lude1 Hm.. this is what I got when I did ∫ dy/y = ∫ k dt lny = kt y = e^kt Where does the 2 come from? I guessed it would have to be from a square, but when you do the integral of a number (since k is a nonzero constant) you multiply the constant with a variable with an exponent of 1, not 2.
you'd get

ln y = kt+A such that y=ekt+A=Bekt where B is a constant. So B can be any number, such as '2'.
 P: 34 Ah, I forgot about that. Thanks!
Math
Emeritus
Thanks
PF Gold
P: 39,564
 Quote by lude1 1. The problem statement, all variables and given/known data If dy/dt = ky and k is a nonzero constant, than y could be a. 2e^kty
If this means y=-2e^(kty) then its derivative is dy/dt= 2ke^(kty)(ky+ kt(dy/dt)).
dy/dt(1- 2kt e^(kty)= 2k^y e^(kty). No, that differential equation is not satisfied.

 b. 2e^kt
If y= 2e^(kt), then dy/dt= 2(ke^(kt))= k(2e^(kt))= ky and satisfies the equation.
(I am assuming the parentheses combine these as you wanted- that you did not intend 2(e^k)t.)

 c. e^kt + 3
If y= e^(kt)+ 3, then y'= ke^(kt) which is not ky= k(e^(kt)+ 3)

 d. kty + 5
If y= kty+ 5, then dy/dt= ky+ kt dy/dt so dy/dt(1- kt)= ky and dy/dt= (k/(1-kt) y. That does not satisfy the given equation.

 e. .5ky^2 + .5
If y= .5y^2+ .5, then dy/dt= 1.0 y dy/dt so dy/dt(1- y)= 0. Either dy/dt= 0 or y= 1. That does not satisfy the given equation.

 Correct answer is b. 2e^kt 2. Relevant equations 3. The attempt at a solution I don't really know what I should be doing. I have a feeling I have to work backwards, therefore do the integral of ky? But if that is the case, I have no idea how e^kt is created through that integral. 2e^kt reminded me of Ce^kt, but that doesn't have anything to do with this problem either. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution

 Related Discussions High Energy, Nuclear, Particle Physics 31 Precalculus Mathematics Homework 1 Special & General Relativity 8 Calculus & Beyond Homework 1 Electrical Engineering 2